I got 68.8403533 percentile in jee main paper 2 how much rank will I get?? Anyone??
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Dear aspirant,
Your rank will be nearly (100-68.84)*145386/100=45302 based on previous year trends.And chances are less to be qualified to appear for jee advanced exam as experts predicted cutoff to be nearly 85 percentile for general candidate.
Good luck
Hello Nidish,
Since approximately around 145386 students appeared for JEE B. Arch exam so the rank can be predicted from the given formula -
(100- your total percentile score) * 1,45,386/ 100.
So, accordingly, your rank is somewhere around 45301 - 45302 when calculated with the percentile given by you. But, this is a rough evaluation so it is best that you wait till the final rank list is announced by NTA. Along with the result of JEE Main Paper-2 April session, authorities will also release the rank list.
For more details, you can go through the following link -
Https://engineering.careers360.com/articles/jee-main-paper-2-result
Good luck!
Hello Nidish,
Since approximately around 145386 students appeared for JEE B. Arch exam so the rank can be predicted from the given formula -
(100- your total percentile score) * 1,45,386/ 100.
So, accordingly, your rank is somewhere around 4529 - 4530 when calculated with the percentile given by you. But, this is a rough evaluation so it is best that you wait till the final rank list is announced by NTA. Along with the result of JEE Main Paper-2 April session, authorities will also release the rank list.
For more details, you can go through the following link -
https://engineering.careers360.com/articles/jee-main-paper-2-result
Good luck!
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